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As far as I know, a Newtonian fluid follows the equation:

$\frac{du}{dy}=\frac{\tau}{\mu}$ (where $u$ is $x$-axis speed and $y$ the height from the contact surface fluid-solid.)

This leads to the conclusion that:

$u=\frac{\tau}{\mu} y ,$

which is clearly a linear speed distribution. However, I am not sure this is a 100% accurate.

So, the question is:

If a fluid has a non-linear speed distribution does this always mean it is a non-Newtonian fluid? Or could this mean it is not a steady flow? I can't come up with more possibilities. Are there any?

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  • $\begingroup$ Hmmm... I think this is probably only true near the boundary either way. If you think about two surfaces producing a shear of a newtonian fluid, and the relative motion of the two surfaces is sufficiently fast, then the Reynolds number may become large somewhere, even though it's small near the surfaces. Once we are leaving a laminar flow regime, the simple linear velocity law probably has to be modified (if there isn't outright turbulence). $\endgroup$ – CuriousOne Oct 11 '14 at 16:11
  • $\begingroup$ I think that equation is a little backwards (should be $\tau=\mu(du/dy)$), but your integration assumes that $\tau$ is a constant, which may not necessarily be the case. $\endgroup$ – Kyle Kanos Oct 11 '14 at 16:40
  • $\begingroup$ Your point is totally accurate. Edited now. However, if $\tau$ isn't constant through time the equation gets unaltered as it is not time-dependant. It just depends on the actual situation, or at least this is how I see it. However, varying $\tau$ through space (through $y$ in a two dimensional problem) will probably change the speed distribution. But is that possible?--No idea. $\endgroup$ – Ioannes Oct 11 '14 at 16:53
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The definition that you are using is not the most general. If you insist on applying the way you do, it only applies for a uniform flow in between two counter-mooving walls. Then, the velocity profile is indeed linear.

A Newtonian fluid is defined by the approximation that local stress (or drag) is proportional to local strain. I would write your equation as,

$$ \tau_{ij}(\vec{y}) = \mu \left(\frac{\partial u_i}{\partial y_j}(\vec{y}) + \frac{\partial u_j}{\partial y_i}(\vec{y}) \right)\, . \hspace{2cm} (*)$$

$\vec{u}(\vec{y})$ is the local velocity field. Note that $\tau_{ij}(\vec{y})$ can be a complicated function of $\vec{y}$. The left hand side of equation $(*)$ is the stress on the fluid at position $\vec{y}$. It has two indices because strain affects each component of $\vec{u}(\vec{y})$ in each direction differently. Under stress alone the local velocity field will change according to,

$$ \partial_t u_i(\vec{y}) = \sum_j \partial_{y_j} \tau_{ij}(\vec{y}) \, . \hspace{2cm} (**)$$

The general case will of course involve the full Navier-Stokes equation with this term included. Note that $(**)$ is just a definition of $\tau_{ij}(\vec{y})$ which applies in the most general cases. It just states that the local velocity field will be affected by its neighbours.

The right-hand side of $(*)$ represents the the Newtonian approximation. It is natural to assume that if the fluid is uniform (i.e. its spatial derivatives vanish) it will not be under strain. If it is not uniform (i.e. when different parts of the fluid move at different speeds) however there will be a strain and $\tau_{ij}(\vec{y})$ may be a complicated function of the derivatives of $\vec{u}(\vec{y})$. Assuming that this function can be Taylor expanded, the leading term (for small inhomogeneities, $\partial_{y_i}u_j(\vec{y})$ small) is given by the right hand side of equation $(*)$.

Note that your equation follows from $(*)$ if the fluid only changes in one direction, $\vec{u}(x,y,z) = \vec{u}(x)$.

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